laplacetransform 3e^{2t}sin(4t)

解答

拉普拉斯变换

3 e^{2 t} sin (4 t)

\frac{12}{( s - 2 ) ^{2} + 16}

求解步骤

L {3 e^{2 t} sin (4 t)}

利用拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 3 L {e^{2 t} sin (4 t)}

L {e^{2 t} sin (4 t)} : \frac{4}{( s - 2 ) ^{2} + 16}

= 3 \cdot \frac{4}{( s - 2 ) ^{2} + 16}

整理

3 \frac{4}{( s - 2 ) ^{2} + 16} : \frac{12}{( s - 2 ) ^{2} + 16}

= \frac{12}{( s - 2 ) ^{2} + 16}

\frac{\partial}{\partial β} (sin (α) cos (β))

\frac{\partial}{\partial y} (x^{y^{z}})

\frac{d}{d x} (ln (\frac{2 x + 1}{x ^{2} - 2}))

\frac{\partial}{\partial x} (4 x^{0.5} + 2 x^{0.25})

\int 2 x e^{4 x} d x